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But if I wanted to find the probability of 3 players in a 9 handed game getting the above pairs, it would *not* be:
9! = 362880
362880*(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) =~ 1413785 to 1
Right? Instead I have to figure out how many ways 3 hands can be distributed amongst 9 players? Or is 3! correct regardless of the number of players in the hand, since we are only concerned about the 3 players who get one of the 3 pairs?
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You use 3! for the probability of 3 specific players having these hands, and if you want the probability that at least 3 players out of 9 players have these hands, as a first approximation you would multiply the probability for 3 specific players by C(9,3) ways to choose the 3 players, like this:
C(9,3)*3!*(4/52)*(3/51)*(4/50)*(3/49)*(4/48)*(3/47) =~ 16,830-to-1.
This is close enough for all practical purposes, but it over counts the hands where more than 3 players have these hands. Since this is low probability, the approximation is very good. For an exact solution, we would have to correct for the over counting by computing successive terms using the
inclusion-exclusion principle.